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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>If <span class="process-math">\(\lambda&gt;0\text{,}\)</span> we write <span class="process-math">\(\lambda = k^2\)</span> where <span class="process-math">\(k &gt; 0\text{,}\)</span> and the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(x)=c_1\cos {kx} + c_2\sin {kx},\quad y'(x)=-kc_1\sin kx + kc_2\cos kx.
\end{equation*}
</div>
<p class="continuation">We now check the boundary conditions.</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\left\{\begin{array}{l} y(0)=c_1=0\\
y'(L)=kc_2\cos kL=0\end{array}\right.\Rightarrow \left\{\begin{array}{l} c_1=0\\
c_2\cos kL=0\end{array}\right.
\end{equation*}
</div>
<p class="continuation">which implies</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
kL=(n-\frac{1}{2})\pi,\quad \Rightarrow \quad k=\frac{\pi(n-\frac{1}{2})}{L},~~n=1,2,3,\cdots
\end{equation*}
</div>
<p class="continuation">We get the eigenvalues <span class="process-math">\(\lambda_n\)</span> and the corresponding eigenfunction <span class="process-math">\(y_n\)</span> as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\lambda_n=\left(\frac{\pi(n-\frac{1}{2})}{L}\right)^2,~y_n=\sin \frac{\pi(n-\frac{1}{2})x}{L},~n=1,2,3,\cdots
\end{equation*}
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<span class="incontext"><a href="sec7_2.html#p-333" class="internal">in-context</a></span>
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